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1

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

If the transformation $$z = \log \tan {x \over 2}$$ reduces the differential equation

$${{{d^2}y} \over {d{x^2}}} + \cot x{{dy} \over {dx}} + 4y\cos e{c^2}x = 0$$ into the form $${{{d^2}y} \over {d{z^2}}} + ky = 0$$ then k is equal to

A
$$-$$4
B
4
C
2
D
$$-$$2

$${{{d^2}y} \over {d{x^2}}} + \cot x{{dy} \over {dx}} + 4y\cos e{c^2}x = 0$$ অবকল সমীকরণটি স্বাধীন চলরাশি x, $$z = \log \tan {x \over 2}$$ এর দ্বারা z-এ রূপান্তরিত হলে সমীকরণটি হয় $${{{d^2}y} \over {d{z^2}}} + ky = 0$$ । সেক্ষেত্রে k এর মান হবে

A
$$-$$4
B
4
C
2
D
$$-$$2
2

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

The solution of

$$\cos y{{dy} \over {dx}} = {e^{x + \sin y}} + {x^2}{e^{\sin y}}$$ is $$f(x) + {e^{ - \sin y}} = C$$ (C is arbitrary real constant) where f(x) is equal to

A
$${e^x} + {1 \over 2}{x^3}$$
B
$${e^{ - x}} + {1 \over 3}{x^3}$$
C
$${e^{ - x}} + {1 \over 2}{x^3}$$
D
$${e^x} + {1 \over 3}{x^3}$$

$$\cos y{{dy} \over {dx}} = {e^{x + \sin y}} + {x^2}{e^{\sin y}}$$ এর সমাধান হল $$f(x) + {e^{ - \sin y}} = C$$ (C হল যদৃচ্ছ বাস্তব ধ্রুবক) । সেক্ষেত্রে f(x) হবে

A
$${e^x} + {1 \over 2}{x^3}$$
B
$${e^{ - x}} + {1 \over 3}{x^3}$$
C
$${e^{ - x}} + {1 \over 2}{x^3}$$
D
$${e^x} + {1 \over 3}{x^3}$$
3

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
The differential of $$f(x) = {\log _e}(1 + {e^{10x}}) - {\tan ^{ - 1}}({e^{5x}})$$ at x = 0 and for dx = 0.2 is
A
0.5
B
0.3
C
$$-$$ 0.2
D
$$-$$ 0.5

Explanation

We have,

$$f(x) = {\log _e}(1 + {e^{10x}}) - {\tan ^{ - 1}}({e^{5x}})$$

$$ \Rightarrow f'(x) = {{10{e^{10x}}} \over {1 + {e^{10x}}}} - {{5{e^{5x}}} \over {1 + {e^{10x}}}}$$

$$ \Rightarrow f'(x) = {{10{e^{10x}} - 5{e^{5x}}} \over {1 + {e^{10x}}}}$$

$$ \Rightarrow f'(0) = {{10 - 5} \over {1 + 1}} = {5 \over 2}$$

We know that,

$$f(x + \Delta x) - f(x) = f'(x)dx$$

$$ \Rightarrow f(x + \Delta x) - f(x) = {5 \over 2} \times 0.2$$ [$$\because$$ dx = 0.2]

$$ \Rightarrow \Delta f(x) = 0.5$$
$$f(x) = {\log _e}(1 + {e^{10x}}) - {\tan ^{ - 1}}({e^{5x}})$$ -এর x = 0 বিন্দুতে dx = 0.2 -এর জন্য অন্তরকলক হবে
A
0.5
B
0.3
C
$$-$$ 0.2
D
$$-$$ 0.5

Explanation

আমাদের কাছে,

$$f(x) = {\log _e}(1 + {e^{10x}}) - {\tan ^{ - 1}}({e^{5x}})$$

$$ \Rightarrow f'(x) = {{10{e^{10x}}} \over {1 + {e^{10x}}}} - {{5{e^{5x}}} \over {1 + {e^{10x}}}}$$

$$ \Rightarrow f'(x) = {{10{e^{10x}} - 5{e^{5x}}} \over {1 + {e^{10x}}}}$$

$$ \Rightarrow f'(0) = {{10 - 5} \over {1 + 1}} = {5 \over 2}$$

আমরা জানি যে,

$$f(x + \Delta x) - f(x) = f'(x)dx$$

$$ \Rightarrow f(x + \Delta x) - f(x) = {5 \over 2} \times 0.2$$ [$$\because$$ dx = 0.2]

$$ \Rightarrow \Delta f(x) = 0.5$$
4

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
If $$x{{dy} \over {dx}} + y = {{xf(xy)} \over {f'(xy)'}}$$, then | f(xy) | is equal to (where k is an arbitrary positive constant).
A
$$k{e^{{x^2}/2}}$$
B
$$k{e^{{y^2}/2}}$$
C
$$k{e^{{x^2}}}$$
D
$$k{e^{{y^2}}}$$

Explanation

Given,

$$x{{dy} \over {dx}} + y = {{xf(xy)} \over {f'(xy)}}$$ .... (1)

Let, $$xy = v$$

$$ \Rightarrow x{{dy} \over {dx}} + y\,.\,(1) = {{dv} \over {dx}}$$ ..... (2)

From (1) and (2), we get,

$${{dv} \over {dx}} = {{xf(xy)} \over {f'(xy)}}$$

$$ \Rightarrow {{dv} \over {dx}} = {{xf(v)} \over {f'(v)}}$$

$$ \Rightarrow {{f'(v)} \over {f'(v)}}dv = xdx$$

Now, integrating both side, we get

$$\int {{{f'(v)} \over {f(v)}}dv = \int {xdx} } $$

$$ \Rightarrow \log |f(v)| = {{{x^2}} \over 2} + \log K$$ [log K = constant]

$$ \Rightarrow \log |f(xy) - \log K = {{{x^2}} \over 2}$$

$$ \Rightarrow \log {{|f(xy)|} \over K} = {{{x^2}} \over 2}$$

$$ \Rightarrow {{|f(xy)|} \over K} = {e^{{{{x^2}} \over 2}}}$$

$$ \Rightarrow |f(xy)| = K{e^{{{{x^2}} \over 2}}}$$

যদি $$x{{dy} \over {dx}} + y = {{xf(xy)} \over {f'(xy)'}}$$ হয়, তবে | f(xy) | হবে (যেখানে k হল যদৃচছ ধনাত্বক ধ্রুবক।)
A
$$k{e^{{x^2}/2}}$$
B
$$k{e^{{y^2}/2}}$$
C
$$k{e^{{x^2}}}$$
D
$$k{e^{{y^2}}}$$

Explanation

দেওয়া, $$x{{dy} \over {dx}} + y = {{xf(xy)} \over {f'(xy)'}}$$

$$ \Rightarrow {d \over {dx}}(xy) = x{{f(xy)} \over {f'(xy)}}$$

$$ \Rightarrow {{f'(xy)} \over {f(xy)}}d(xy) = x\,dx$$

$$ \Rightarrow \int {{{f'(xy)} \over {f(xy)}}d(xy) = \int {x\,dx} } $$

$$ \Rightarrow \log \left| {f(xy)} \right| = {{{x^2}} \over 2} + C$$

$$ \Rightarrow \left| {f(xy)} \right| = {e^{{{{x^2}} \over 2}}} + C = k{e^{{{{x^2}} \over 2}}}$$

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